Tharrmashastha V
• Member for 2 years, 2 months
• Last seen more than a month ago
• New Delhi, Delhi, India

In your code, circuits is an array of array of quantum circuits and each element of circuits is an array of quantum circuits. In the code, you are using a for loop to run each element of the circuits ...

Say you have two bases $B_1$ and $B_2$. Let $U$ be the unitary that transforms the orthonormal basis vectors in $B_1$ to the basis vectors in $B_2$. So, it is obvious that $U^{\dagger}$ will be the ...

If you recall from the analysis of Grover's algorithm, if the success probability of obtaining an item of interest is $\sin^2(\theta)$ just before a Grover iteration, then just after one iteration of ...

Contrary to what you have mentioned in the question, when you use a $u_3$ gate, Qiskit does not break it up into multiple gates. In fact $u_3$ gate is one of the basis gates for all Qiskit backends ...

Short Answer: There is no universal gate that can entangle the output state for any given ? and ??? operators in the circuit given in the question. Long Answer: Say you have the entangle state $|\psi\... View answer Accepted answer 3 votes We know that operators are applied on qubits. But sometimes it does not make sense to ask the state on which we apply an operator, say$U$. For instance, take the two-qubit entangled state$|\psi\...

Measuring in the 'Computational Basis' always means to measure in the $\{|0\rangle,|1\rangle\}^n$ basis for $n$ qubits. As for your second question, I feel that you misinterpreted the author. It is ...

I assume that you have a qubit register $q$ and given that the state of $q$ is $|\psi_i\rangle$ you want to apply $U_i$ to $q$ for $i=0,1,2$. If this is what you wish to do, then unfortunately if the ...

To add to the circuit by @Craig, the reason why papers tend to mention that some gate is obtainable by $\sqrt{swap}$ upto a few single qubit gates is because the set that contains the $\sqrt{swap}$ ...

If I am correct, I suppose you are talking about the Qiskit Challenge 2020. A possible reason why your circuit is being graded wrong is because the question asks you to construct the circuit for full ...

It is correct. Since you have $R_1(\theta) = \begin{bmatrix}1 & 0\\ 0 & e^{i\theta}\end{bmatrix}$ and $R_z(\theta) = \begin{bmatrix}e^{-i\theta/2} & 0\\ 0 & e^{i\theta/2}\end{bmatrix}$ ...
Quantum phase estimation does not have anything to do with $\theta$. I feel that you are confusing phase estimation with the implementation of HHL as given in the paper https://arxiv.org/abs/1804....